[tex]\huge\star \: \sf{ Question}[/tex]
The length of a rectangle is greater than the breadth by 3cm . If the length is increased by 9cm and the breadth is reduced by 5cm , the area remains the same. Find the dimensions of the rectangle ?
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[tex]\huge\star \: \sf{ Question}[/tex]
The length of a rectangle is greater than the breadth by 3cm . If the length is increased by 9cm and the breadth is reduced by 5cm , the area remains the same. Find the dimensions of the rectangle ?
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Let the breadth be x cm.
Then the length will be (x+3) cm.
Area of the rectangle = x(x+3) cm^2. (1)
Given that length is increased by 9cm = x +3+9 = x + 12 cm.
Given that breadth is reduced by 5cm = x - 5 cm.
Area of the rectangle = (x+12)(x-5). (2)
On solving (1) and (2), we get
x(x+3) = (x+12)(x-5)
x^2 + 3x = x^2 + 7x - 60
4x = 60
x = 15.
Breadth = 15cm.
Length = 15 + 3 = 18 cm.
Step-by-step explanation:
[tex] \sf \: Solution :[/tex]
Let the breadth be x cm
so length will be (x + 3) cm
We know that
[tex] \sf ☞ \: Area \: = Length \times Breadth[/tex]
[tex] \sf \: =x \times (x+3) \\ \sf =(x^ 2 +3x)cm^ 2[/tex]
Given that length is increased by 9 cm=
(x+3+9) = (x+12) cm
Breadth is seduced by 5 cm=
(x-5)cm
[tex] \sf \: Area = l \times b \\ \sf =(x+12) \times (x-5) \\ \sf =x(x-5)12(a-5) \\ \sf =x^ 2 -5x+12x-60 \\ \sf =x^ 2 +7x-60[/tex]
According to the question :
[tex]= \sf \: x^ 2 +3x =x^ 2 +7x-60 \\ \sf = \cancel{x^ 2} - \cancel{ x^ 2} +3x-7x=-60 \\ \sf = 4x \: = 60 \\ \sf = x = \cancel{\frac{60}{4} }[/tex]
[tex] \sf \red{ Breath = 15 cm} \\ \sf \red{Length = (15+3) = 18 cm }[/tex]
Hope it'll help you friend
have a nice day :-D